(20+2x)(16+2x)=1120

Solution for (20+2x)(16+2x)=1120 equation:

(20+2x)(16+2x)=1120

We move all terms to the left:

(20+2x)(16+2x)-(1120)=0

We add all the numbers together, and all the variables

(2x+20)(2x+16)-1120=0

We multiply parentheses ..

(+4x^2+32x+40x+320)-1120=0

We get rid of parentheses

4x^2+32x+40x+320-1120=0

We add all the numbers together, and all the variables

4x^2+72x-800=0

a = 4; b = 72; c = -800;
Δ = b2-4ac
Δ = 722-4·4·(-800)
Δ = 17984
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{17984}=\sqrt{64*281}=\sqrt{64}*\sqrt{281}=8\sqrt{281}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-8\sqrt{281}}{2*4}=\frac{-72-8\sqrt{281}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+8\sqrt{281}}{2*4}=\frac{-72+8\sqrt{281}}{8}
$